Estimate the distance (in nanometers) between molecules of water vapor at \(100^{\circ} \mathrm{C}\) and 1.0 atm. Assume ideal behavior. Repeat the calculation for liquid water at \(100^{\circ} \mathrm{C}\), given that the density of water is \(0.96 \mathrm{~g} / \mathrm{cm}^{3}\) at that temperature. Comment on your results. (Assume water molecule to be a sphere with a diameter of \(0.3 \mathrm{nm} .\) ) (Hint: First calculate the number density of water molecules. Next, convert the number density to linear density, that is, number of molecules in one direction.)

The ideal gas law is a fundamental principle in chemistry and physics, represented by the equation \(PV = nRT\). It correlates the pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), and temperature of an ideal gas to the universal gas constant (\(R\)), providing insights into the behavior of gases under varying conditions.

In context with the exercise, the ideal gas law is employed to calculate the number density, which is the number of gas molecules per unit volume. By knowing the pressure, temperature, and the gas constant in appropriate units, we can derive the number of moles of water vapor. Then, by dividing the number of molecules by Avogadro's number, we obtain the number density in terms of molecules per cubic meter.

This equation assumes that gas molecules do not attract or repel each other and that they occupy no volume, making it an idealization that works well under many common conditions, such as the scenario described in the exercise.

Number density is defined as the number of particles per unit volume, which gives us a measure of how compactly packed the particles are in a given space. It is expressed as \(n/V\) in the ideal gas equation, where \(n\) is the number of molecules and \(V\) is the volume in cubic meters.

In the exercise, number density is central to understanding the molecular distribution in water vapor at a given temperature and pressure. To acquire the number density, we use the ideal gas law to find the amount of substance in moles, which can then be converted to a molecular count by Avogadro's number, providing an insight into how many water molecules there are per cubic meter.

This figure is then used to extract the average distance between the molecules. It's crucial because it represents how the gas molecules are dispersed and gives a sense of their interactions at a specific condition.

Linear density is the count of particles along a single line, or in other words, the number of particles per unit length. Once we have the number density from the previous step, we can calculate the linear density by taking the cube root of the number density.

In the exercise, this calculation is key to finding out the average distance between individual molecules of water in both vapor and liquid states. By assuming uniform distribution, we can infer that the cube root of the number of molecules in a cubic meter gives us the number of molecules along a single meter. Dividing 1 by this linear density grants us the average molecular distance in the material, which in this case, provides comparisons between states of matter.

Avogadro's number, \(6.022 \times 10^{23}\), is the number of atoms, ions, or molecules in one mole of substance. It serves as a bridge between the macroscopic world we experience and the microscopic world of atoms and molecules.

In our exercise, we use Avogadro's number to convert moles, determined using the ideal gas law, into the actual count of water molecules. Since the number density requires an actual count of molecules and not moles, Avogadro's number is crucial in this conversion process. Understanding how to use Avogadro's number to navigate between moles and number of particles empowers students to tackle various problems regarding molecular quantities and concentrations.

The state of matter of a substance is determined by its physical form, which is a result of the kinetic energy of its particles and intermolecular forces. Common states include solid, liquid, gas, and plasma. In the exercise, we are comparing water in its gas (vapor) and liquid forms at the same temperature.

The molecules in the liquid state are more densely packed, hence they have a shorter distance between them compared to the gaseous state. This contrast is illustrated by the distance calculations from the molecular count. The exercise highlights how the state of matter affects the distribution and spacing of molecules, reinforcing the idea that gases have much lower densities and larger spaces between molecules than liquids.